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Irina-Kira [14]
3 years ago
8

Create a linear function that has a rate of change of -3 and a solution at (5,2)

Mathematics
1 answer:
jeyben [28]3 years ago
8 0

Answer:

y = - 3x + 17

Step-by-step explanation:

Note that slope is a measure of the rate of change of a linear function.

Express the function in slope- intercept form

y = mx + c ( m is the slope and c the y- intercept )

here m = - 3, thus

y = - 3x + c ← is the partial equation

To find c substitute (5, 2) into the partial function

2 = - 15 + c ⇒ c = 2 + 15 = 17

y = - 3x + 17 ← is the linear function

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Graph the function f(x) = -x-8x-13
jekas [21]

Answer:

the answer is there...

5 0
3 years ago
What is the answer to this problom
Liula [17]
The answer would be 405 because
    
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    6|2430                     sorry for long example.
       24
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7 0
3 years ago
List all possible rational roots. Then use synthetic division to confirm which rational roots exist:
Kisachek [45]

Answer:

\boxed{(1) \, x = \, \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10; (2) \, x = -2}

Step-by-step explanation:

2x³+ 6x² - x - 10 = 0

(1) Possible roots

The Rational Roots Theorem states that, if a polynomial has any rational roots, they will have the form p/q, where p is a factor of the constant term  and q is a factor of the leading coefficient.

\text{Possible rational root} = \dfrac{ p }{ q } = \dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}

In your function, the constant term is -10 and the leading coefficient is 2, so

\text{Possible root} = \dfrac{\text{factor of 10}}{\text{factor of 2}}

Factors of 10 = ±1, ±2, ±5, ±10

Factors of 2 = ±1, ±2

\text{Possible roots are } \large \boxed{\mathbf{x = \pm \dfrac{1}{2}, \pm 1, \pm2, \pm \dfrac{5}{2}, \pm 5, \pm 10}}

(2) Synthetic division

Rather than work through all 12 possibilities, I will do one that works.

\begin{array}{r|rrrr}-2 & 2 & 6 & -1 & -10\\& & -4& -4 & 10\\& 2 & 2& -5 & 0\\\end{array}

So, x = -2 is a root, and the quotient is 2x² + 2x - 5.

(3) Check for other rational roots

2x² + 2x - 5 = 0

D = b² - 4ac =2²- 4(2)(-5) = 4 + 40 = 44

√44 = 2√11, which is irrational.

Since irrational roots come in pairs, the cubic equation has two real, irrational roots and one rational root at x = -2.

6 0
4 years ago
6A. make a box and whiskers plot for the following data label the extrems,quartites class score 40,30,75,82,90,25,80,46
Aliun [14]
<span> 25  30  35  40  46  60.5  75  80  81  82  90
min      Q1           Q2/med          Q3      max
81-35=46=Interquartile range

</span><span>Is that all the information you wanted?</span>
5 0
3 years ago
1-12 I don’t know if you can please help me thanks
katovenus [111]

1 24

2 .41

3 180

4  2000

5   46

6  .18

7   60.5666

8   6.5

9   128

10   .8

11    3.628739

12   4000


pretty sure juh looked it up hope it helps!

4 0
3 years ago
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